Determine set {$w=\frac{2}{3-z}:z=2+iy$, $y \in \mathbb{R}$} in complex plane.
I've tried with putting $z$ in denominator and rationalizing.
$w=\frac{2}{1-iy}\cdot\frac{1+iy}{1+iy}=\frac{2+2iy}{1+y^2}=\frac{2}{1+y^2}(1+iy)$,
The solution says that this is circle with center in 1, radius 1, and origin is excluded. I just can't get it. Can someone give me a hint?
Let $w=u+iv$, $u$ and $v$ real. Then $u+iv=\frac 2{1-iy}=\frac {2(1+iy)} {1+y^{2}}$ . Hence $u=\frac 2 {1+y^{2}}$ and $v=\frac {2y} {1+y^{2}}$. Thus $$(u-1)^{2}+v^{2}=\frac {(1-y^{2})^{2}}{(1+y^{2})^{2}}+\frac {4y^{2}} {(1+y^{2})^{2}}$$ $$=\frac {(1+y^{2})^{2}} {(1+y^{2})^{2}} =1.$$ This equation represents the circle with center $(1,0)$ and radius $1$. [As $y$ varies over $\mathbb R$ all points in this circle are covered. Can you see why?].
